Question:

Step 1: The greatest integer function $[x]$ is discontinuous at every integer $n \in \mathbb{Z}$. Specifically, at an integer $n$, $\lim_{x \to n^-} [x] = n-1$ and $\lim_{x \to n^+} [x] = n$. Step 2: The function $\cos\left(\frac{2x-1}{2\pi}\right)$ is a composition of continuous functions ($2x-1$, division by $2\pi$, and $\cos(u)$), and is therefore continuous for all $x \in \mathbb{R}$. Step 3: Consider the continuity of $f(x)$ at integer points $n \in \mathbb{Z}$. The left-hand limit at $x=n$ is: $$ \lim_{x \to n^-} f(x) = \lim_{x \to n^-} [x] \cos\left(\frac{2x-1}{2\pi}\right) = (n-1) \cos\left(\frac{2n-1}{2\pi}\right) $$ The right-hand limit at $x=n$ is: $$ \lim_{x \to n^+} f(x) = \lim_{x \to n^+} [x] \cos\left(\frac{2x-1}{2\pi}\right) = n \cos\left(\frac{2n-1}{2\pi}\right) $$ For these limits to be equal, we must have $(n-1) \cos\left(\frac{2n-1}{2\pi}\right) = n \cos\left(\frac{2n-1}{2\pi}\right)$. This implies $(n-1-n) \cos\left(\frac{2n-1}{2\pi}\right) = 0$, which simplifies to $-\cos\left(\frac{2n-1}{2\pi}\right) = 0$. For $\cos(\theta) = 0$, $\theta$ must be an odd multiple of $\frac{\pi}{2}$. Thus, we would need $\frac{2n-1}{2\pi} = \frac{k\pi}{2}$ for some odd integer $k$. This leads to $2n-1 = k\pi^2$. Since $2n-1$ is an integer and $k\pi^2$ is an irrational number (as $k$ is a non-zero integer), this equality is impossible. Therefore, $\cos\left(\frac{2n-1}{2\pi}\right) \neq 0$ for any integer $n$. Since $\cos\left(\frac{2n-1}{2\pi}\right) \neq 0$ and $n-1 \neq n$, the left-hand limit and the right-hand limit are not equal at any integer $n$. Thus, $f(x)$ is discontinuous at every integer point. Step 4: Consider the continuity of $f(x)$ at non-integer points $x \notin \mathbb{Z}$. For any $x$ that is not an integer, the greatest integer function $[x]$ is continuous. As established in Step 2, $\cos\left(\frac{2x-1}{2\pi}\right)$ is continuous for all $x \in \mathbb{R}$. Since $f(x)$ is the product of two functions that are continuous at non-integer points, $f(x)$ is continuous at all non-integer points. Conclusion: The function $f(x)$ is discontinuous at all integer points.